method 2

Take log both sides and get
\[\left(x_n+n\right)\ln\left(1+\frac{1}{n}\right)=1\]
Also as \[t=\frac{1}{n};n\to\infty;t\to0\]
Now \[x_n=\frac{1}{\ln\left(1+t\right)}-\frac{1}{t}=\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}\]
\[\lim_{n\to\infty}x_n=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2}\cdot\lim_{t\to0}\frac{1}{\frac{\ln\left(1+t\right)}{t}}=\frac{1}{2}\cdot1\]
(Used certain standard limits which you can solve by series or wikipedia for more methods.)

Comments

The issue with your first method is that when you took limits in the first step, you increased the solution space of \(x_n\) from being an exact value, to being a much larger set. This is why you reach the conclusion that \(x_n\) cannot be determined. As you denoted, you only have forward implication signs, but not backward implication.

Note: It is true that if the limit of \(Y_n\) is \( \frac{1}{2} \) (or any other constant), then the limit of \( \left( 1 + \frac{1}{n} \right) ^{Y_n} =1 \).

The simple english version is that your implications only work one way (as indicated by your forward arrows). Hence, the given condition implies your final conclusion, but not necessarily the other way around.

and claiming that \( x = -1 \) is a solution to the original equation. This is not true, because the implications only work one way. We only have \( x = 1 \Rightarrow x = 1 \text{ or } -1 \) but we do not have \( x = 1 \text{ or } -1 \Rightarrow x = 1 \).

You can see how in the first implication, you increased the solution space from 1 element, to 2 elements.